It is known that the radius of the arc is 441 mm and the arc length is 307 mm. What are the height and length of the arc

It is known that the radius of the arc is 441 mm and the arc length is 307 mm. What are the height and length of the arc

Center angle = 0.307 / (0.441 * 2 * Π) * 360 = 39.898m
Then: Xuanchang = 2 * sin19.95 * 0.441 = 0.3m
Xuangao = 0.441 - (0.15 / 19.95) = 0.028m

What is the geometric center in mathematics?
The geometric center of a circle is the center of a circle, and the geometric center of a rectangle is the intersection of diagonals. How do you judge this? What is the geometric center of any triangle?

The geometric center is the center of gravity (the intersection point of the center line). The geometric center of any triangle is the intersection point of the center line!
When a line divides a figure into two parts of equal area, it is the center line

First simplify and then evaluate, (A's Square / a square - 5A + 6) / (a square + A-2) / (A-3) - (a + 3) / (a square - 4), where a = - 3

The original formula = [a ^ 2 / (a -- 2) (a -- 3)] / [(a + 2) (a -- 1)] / [(a -- 3)] - [(a + 3) / (a + 2) (a -- 2)] = [a ^ 2 / (a -- 2) (a -- 3)] * [(a -- 3) / (a + 2) (a -- 1)] - [(a + 3) / (a + 2) (a -- 2)] = [a ^ 2 / (a + 2) (a -- 1) -[(a + 3) / (a + 2) (a -- 2)] = [a ^ 2 / (a + 2) (a -- 1)] - [(a

It is known that ABC is a real number, the function f (x) = ax6 ^ 2 + BX + C, G (x) = ax + B, when - 1 ≤ x ≤ 1 | f (x) | ≤ 1. It is proved that | C | ≤ 1
2. It is proved that when - 1 ≤ x ≤ 1 | g (x) | ≤ 2 3. Let a > 0, when - 1 ≤ x ≤ 1, the maximum value of G (x) is 2, and f (x)
When someone answers "know", the plus sign is omitted, but even if the space in the middle is "| g (1) | = | a B"|

To prove that | g (1) | = | a + B|

Factorization: (1) 2x (a-b) - (B-A); (2) x (x + 4) + 3; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; (3) x2-2x + 1-y2

(1) Original formula = 2 & nbsp; X & nbsp; (a-b) + (a-b) = (a-b) (2 & nbsp; X + 1); (2) original formula = x2 + 4x + 3 = (x + 1) (x + 3); (3) original formula = (x-1) 2-y2 = (x + Y-1) (x-y-1)

It is known that there are three points a (- 2,3) B (- 1,0) C (0,1) on the image of quadratic function?

The solution consists of a quadratic function with C (0,1)
Let a quadratic function be
y=ax^2+bx+1
It is shown that there are three points a (- 2,3) B (- 1,0) on the image of quadratic function
Then 4a-2b = 2
a-b=-1
The solution is a = 2, B = 3
so
The quadratic function is
y=2x^2+3x+1

2X & # 178; - X-6 = 0 use the collocation method to solve the equation

2x²-x-6＝0
x^2-x/2=3
(x-1/4)^2=17/16
x-1/4=±√17/4
x1=1/4+√17 x2=1/4-√17 .

Find the trajectory equation of the vertex of parabola y = x ^ 2 + (2m + 1) x + m ^ 2-1 (M is a real number)

Y = x ^ 2 + (2m + 1) x + m ^ 2-1 = {x ^ 2 + 2 * [(2m + 1) / 2] x + [(2m + 1) / 2] ^ 2} - [(2m + 1) / 2] ^ 2 + m ^ 2-1 = [x + (2m + 1) / 2] ^ 2 + (2m ^ 2 + M-3 / 4) so the vertex [- (2m + 1) / 2,2m ^ 2 + M-3 / 4] is x = - (2m + 1) / 2, y = 2m ^ 2 + M-3 / 4m = - X-1 / 2 substituted by y = 2m ^ 2 + M-3 / 4Y = 2x ^ 2 + x-3 / 4

Two pursuit and the problem of the application of the equation of one variable (can do it quickly
1. On the overtaking lane, how many seconds does it take for a car with a length of 4 meters and a speed of 100 km / h to overtake a truck with a length of 12 meters and a speed of 80 km / h? (two significant figures are reserved)
2. The distance between the two stations is 448 km. A slow train starts from the station and an express train starts from the station. It runs 60 km and 80 km per hour respectively. The two trains leave at the same time and go in the same direction. If the slow train is ahead, how many hours will the express catch up with the slow train?

1: First, convert 100 km / h to 250 / 9 m / s, 80 km / h to 200 / 9 m / s, then this is the pursuit problem, so the actual speed can be regarded as 50 / 9 m / s, the length is 12 + 4 = 16 m, so the time t = 16 / (50 / 9) = 2.88 s 2: so simple, take the slow train as the reference system, that is, the fast train takes 20 km / h

Given that the distance between point m and X axis and the distance between point m and point F (0,4) are equal, the trajectory equation of point m is obtained

If M (x, y) is set as a moving point, then the distance between M and X axis and the distance between M and f (0, 4) are equal, that is, the trajectory equation of M is y = 18x2 + 2